What is M8.4: Newton's third law and equilibrium?

Draw a separate force diagram for each body, then resolve: in equilibrium the components in two perpendicular directions each sum to zero; in motion the resultant equals ma along the line of motion, with one sign convention throughout.

How many marks are M8.4 questions worth?

M8.4 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has M8.4 appeared in?

We have not yet indexed a verified past-paper appearance for M8.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M8.4

Understand and use Newton's third law; equilibrium of forces on a particle and motion in a straight line; apply to smooth pulleys and connected particles; resolve forces in 2 dimensions; equilibrium of a particle under coplanar forces.

Draft — not yet indexed

Newton's third law and equilibrium

Worked answers and methods for M8.4 on Edexcel A-level Maths 9MA0.

Definition

Specification M8.4: understand and use Newton's third law; equilibrium of forces on a particle and motion in a straight line; application to problems involving smooth pulleys and connected particles; resolving forces in two dimensions; equilibrium of a particle under coplanar forces.

Explanation

Draw a separate force diagram for each body, then resolve: in equilibrium the components in two perpendicular directions each sum to zero; in motion the resultant equals $ma$ along the line of motion, with one sign convention throughout.
For connected particles joined by a light inextensible string over a smooth light pulley, the tension is the same on both sides and both particles have the same magnitude of acceleration, so write $F=ma$ for each particle separately and solve the pair simultaneously;
the string, pulley and modelling assumptions are what justify those two facts and are the answer when the question asks why.
On an inclined plane resolve along and perpendicular to the plane, with $R=mg\cos\alpha$ , never $mg$ , and the weight component $mg\sin\alpha$ down the slope.
Newton's third law pairs act on different bodies, so the force on the pulley is $2T$ downwards. Give forces in newtons with $g=9.8$ , to two or three significant figures.

Two particles connected by a light string over a smooth pulley share one tension magnitude while the string is taut.

Worked example

Two particles

A

and

B

, of masses

5

kg and

3

kg respectively, are connected by a light inextensible string passing over a small smooth fixed pulley. The particles hang freely and the system is released from rest with the string taut. Find the acceleration of the particles and the tension in the string.

(6)

(Total for Question 1 is 6 marks)

Working

1.For $A$ , taking downwards as positive: $5g-T=5a$ .
2.For $B$ , taking upwards as positive: $T-3g=3a$ .
3.Add the equations to eliminate $T$ : $2g=8a$ , so $a=\dfrac{g}{4}=2.45$ m s $^{-2}$ .
4.Substitute back: $T=3g+3a=3(9.8)+3(2.45)=36.75$ N, that is $37$ N to 2 s.f.

Answer: $a=2.45$ m s $^{-2}$ ; $T=36.75$ N ( $37$ N)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Equation of motion for $A$ , with the correct terms: $5g-T=5a$	M1
	$5g-T=5a$	A1
	Equation of motion for $B$ : $T-3g=3a$	M1
	$T-3g=3a$	A1
	$a=2.45$ m s $^{-2}$	A1
	$T=36.75$ N	A1
Notes
M1: Condone sign errors
A1: Or the whole-system equation $2g=8a$ for M1A1
A1: Accept $\frac{g}{4}$ or $2.5$
A1: Accept $37$ N; $\frac{15g}{4}$

(6 marks)

Common mistakes

Don't assign different tension magnitudes to the two ends of one light inextensible string over a smooth pulley.
Don't treat a Newton third-law pair as two forces on the same particle or assume connected-particle tensions without a force diagram.

Mark-scheme wording to use

“0.5g, (1/2)g or 4.9 (N) seen … B1 cao. (must be positive) … This answer must appear in (a) to earn this mark.” — Pearson 9MA0, June 2024 Paper 32, Q1(a)
“M1 for resolving horizontally or an equivalent complete method. A1 for X = Tcos(60). M1 for resolving vertically or an equivalent complete method. A1 for Y = Tcos(30) - 20g as the downward component magnitude” — Pearson 9MA0, Specimen Paper 32, Q4(b)
“B1 for R = mg cos(alpha). M1 for resolving parallel to the plane. A1 for the correct equation -F - mg sin(alpha) = -0.8mg, or a sign-equivalent form” — Pearson 9MA0, Sample Paper 32, Q2(a)

What the examiner wants

Each stage named and the diagram annotated. The June 2024 Paper 32 report says the most successful candidates indicated what they were doing at each stage, resolving horizontally or taking moments about $A$ , and that on the inclined plane those who used $\tan$ , or divided by $\sin$ or $\cos$ instead of writing $R=mg\cos\alpha$ , lost both marks.

Exam tip

Draw a separate force diagram for each particle and distinguish interaction pairs from forces that can balance.

Worked practice

Tier 1 · Easy

A book pushes down on a table with force

P

. State the Newton's third-law partner to this force.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	The table pushes upwards on the book with force $P$	2
Notes
The partner must be the same interaction with the bodies reversed. It is the force of the table on the book, equal in magnitude to $P$ and opposite in direction.

(2 marks)

Tier 2 · Standard

Particles of masses

3\,\text{kg}

and

5\,\text{kg}

hang at the ends of a light inextensible string passing over a smooth pulley. Using

g=9.8\,\text{m s}^{-2}

, find the acceleration of the system and the tension in the string.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Acceleration $2.45\,\text{m s}^{-2}$ , with the $5\,\text{kg}$ particle moving down Tension $36.75\,\text{N}$ , or $36.8\,\text{N}$ to 3 significant figures	5
Notes
Treating both particles as one system, the driving force is $(5-3)g=19.6\,\text{N}$ and the total mass is $8\,\text{kg}$ , so $a=19.6/8=2.45\,\text{m s}^{-2}$ . For the $3\,\text{kg}$ particle moving upward, $T-3g=3a$ , so $T=3(9.8+2.45)=36.75\,\text{N}$ .

(5 marks)

Tier 3 · Hard

4\,\text{kg}

particle on a smooth plane inclined at

30^\circ

is connected by a light inextensible string over a smooth pulley to a freely hanging

3\,\text{kg}

particle. Using

g=9.8\,\text{m s}^{-2}

, find the acceleration of the system and the tension in the string.

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Acceleration $=1.4\,\text{m s}^{-2}$ , with the $3\,\text{kg}$ particle moving down Tension $=25.2\,\text{N}$	7
Notes
The downslope weight component of the $4\,\text{kg}$ particle is $4(9.8)\sin30^\circ=19.6\,\text{N}$ , while the hanging weight is $29.4\,\text{N}$ , so the $3\,\text{kg}$ particle moves down. For the system, $29.4-19.6=7a$ , giving $a=1.4\,\text{m s}^{-2}$ . For the hanging particle, $29.4-T=3(1.4)$ , so $T=25.2\,\text{N}$ .

(7 marks)

Tier 1 · Easy

A tow rope pulls a sledge forwards with force

180\,\text{N}

. State the Newton's third-law partner to this force.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	The sledge pulls the tow rope backwards with force $180\,\text{N}$	2
Notes
The partner force belongs to the same rope-sledge interaction, has equal magnitude and opposite direction, and acts on the other body. It is therefore the $180\,\text{N}$ backward force of the sledge on the rope.

(2 marks)

Tier 2 · Standard

A particle of weight

60\,\text{N}

is held in equilibrium by two light strings. One string extends horizontally to the left from the particle. The other slopes upwards to the right at

45^\circ

to the horizontal. Find the tension in each string.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Tension in the sloping string $=60\sqrt2\,\text{N}$ Tension in the horizontal string $=60\,\text{N}$	4
Notes
Let the sloping tension be $T$ and the horizontal tension be $H$ . Vertical equilibrium gives $T\sin45^\circ=60$ , so $T=60\sqrt2\,\text{N}$ . Horizontal equilibrium then gives $H=T\cos45^\circ=60\,\text{N}$ .

(4 marks)

Tier 3 · Hard

Particles

A

and

B

, of masses

3\,\text{kg}

and

2\,\text{kg}

respectively, lie on two smooth planes inclined at

30^\circ

and

45^\circ

to the horizontal. They are connected by a light inextensible string passing over a smooth pulley at the common top of the planes. The string lies along a line of greatest slope of each plane. The particles are released from rest. Using

g=9.8\,\text{m s}^{-2}

, find the direction of motion, the acceleration and the tension, giving numerical answers to 3 significant figures.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$A$ moves down the $30^\circ$ plane and $B$ moves up the $45^\circ$ plane Acceleration $=0.168\,\text{m s}^{-2}$ Tension $=14.2\,\text{N}$	6
Notes
The downslope weight components are $3g\sin30^\circ=14.7\,\text{N}$ for $A$ and $2g\sin45^\circ=9.8\sqrt2\,\text{N}$ for $B$ . Since $14.7>9.8\sqrt2$ , $A$ moves down its plane. For the system, $14.7-9.8\sqrt2=5a$ , so $a=0.168\,\text{m s}^{-2}$ to 3 significant figures. For $A$ , $14.7-T=3a$ , giving $T=14.2\,\text{N}$ to 3 significant figures.

(6 marks)

Tier 2 · Standard

Blocks

A

and

B

, of masses

4\,\text{kg}

and

6\,\text{kg}

respectively, are in contact on a smooth horizontal plane. A horizontal force of

30\,\text{N}

is applied to

A

towards

B

. Find the acceleration of the blocks and the force exerted by

A

B

. State the Newton's third-law partner to this contact force.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Acceleration $=3\,\text{m s}^{-2}$ Force exerted by $A$ on $B=18\,\text{N}$ in the direction of motion The partner is the $18\,\text{N}$ force exerted by $B$ on $A$ in the opposite direction	5
Notes
Treating both blocks as one system, $30=(4+6)a$ , so $a=3\,\text{m s}^{-2}$ . For block $B$ , the only horizontal force is the contact force from $A$ , so it is $6(3)=18\,\text{N}$ in the direction of motion. By Newton's third law, $B$ exerts an $18\,\text{N}$ force on $A$ in the opposite direction.

(5 marks)

Tier 3 · Hard

A particle

A

of mass

5\,\text{kg}

rests on a smooth horizontal table and is connected to a hanging particle

B

of mass

3\,\text{kg}

by a light inextensible string passing over a small smooth pulley at the edge of the table. The string is horizontal between

A

and the pulley. Starting from rest, the particles move with the same acceleration, the string being taut. Take

g=9.8\,\text{m s}^{-2}

. Find their acceleration and the tension in the string. Find their speed after

B

has descended

1.5\,\text{m}

, giving the speed to 3 significant figures. Use unrounded values in your working.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Acceleration $=3.675\,\text{m s}^{-2}$ Tension $=18.375\,\text{N}$ Speed $=3.32\,\text{m s}^{-1}$	6
Notes
Let the acceleration be $a$ and the tension be $T$ . For $A$ , $T=5a$ . For $B$ , taking downwards as positive, $3g-T=3a$ . Hence $29.4=8a$ , so $a=3.675\,\text{m s}^{-2}$ and $T=18.375\,\text{N}$ . Using $v^2=u^2+2as$ with $u=0$ and $s=1.5$ , $v=\sqrt{2(3.675)(1.5)}=3.320\ldots\,\text{m s}^{-1}$ , which is $3.32\,\text{m s}^{-1}$ to 3 significant figures.

(6 marks)

Tier 3 · Hard

Particles

A

B

and

C

, of masses

2\,\text{kg}

3\,\text{kg}

and

5\,\text{kg}

respectively, lie in that order on a smooth horizontal plane. Adjacent particles are connected by separate light inextensible strings. A horizontal force of

50\,\text{N}

is applied to

C

away from

B

, and both strings remain taut. Find the acceleration of the particles and the tension in each string. State the Newton's third-law partner to the force exerted by the string joining

B

and

C

on particle

C

. Give all numerical answers exactly.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Acceleration $=5\,\text{m s}^{-2}$ Tension between $A$ and $B=10\,\text{N}$ Tension between $B$ and $C=25\,\text{N}$ The partner is the equal and opposite force exerted by $C$ on that string	6
Notes
For the complete system, $50=(2+3+5)a$ , so $a=5\,\text{m s}^{-2}$ . For $A$ , the first tension is the only horizontal force, so $T_1=2(5)=10\,\text{N}$ . The second tension accelerates $A$ and $B$ together, so $T_2=(2+3)(5)=25\,\text{N}$ . The third-law partner to the string's force on $C$ is the force of $C$ on the same string, equal in magnitude and opposite in direction.

(6 marks)

Q10

Tier 3 · Hard

10.

A particle

B

of mass

4\,\text{kg}

lies on a smooth horizontal table. It is connected on its left to a hanging particle

A

of mass

2\,\text{kg}

by one light inextensible string over a smooth pulley, and on its right to a hanging particle

C

of mass

6\,\text{kg}

by a separate light inextensible string over a second smooth pulley. The horizontal parts of both strings are collinear, and both strings remain taut. The system is initially stationary and is then set free. Use

g=9.8\,\text{m s}^{-2}

. Find the direction and exact magnitude of the acceleration, and find the tension in each string. Give all numerical answers exactly.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$C$ moves down, $B$ moves right and $A$ moves up Acceleration $=\dfrac{49}{15}\,\text{m s}^{-2}$ Left-string tension $=\dfrac{392}{15}\,\text{N}$ Right-string tension $=\dfrac{196}{5}\,\text{N}$	7
Notes
The greater hanging weight is that of $C$ , so take $C$ down, $B$ right and $A$ up as positive. For the whole system the driving force is $(6-2)g=39.2\,\text{N}$ and the total mass is $12\,\text{kg}$ , giving $a=39.2/12=49/15\,\text{m s}^{-2}$ . For $A$ , $T_L-2g=2a$ , so $T_L=19.6+98/15=392/15\,\text{N}$ . For $C$ , $6g-T_R=6a$ , so $T_R=58.8-294/15=196/5\,\text{N}$ .

(7 marks)

Verified exam appearances

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Other points in M8 Forces and Newton's laws

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